Problem: Is ${82971}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {82971}= &&{8}\cdot10000+ \\&&{2}\cdot1000+ \\&&{9}\cdot100+ \\&&{7}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {82971}= &&{8}(9999+1)+ \\&&{2}(999+1)+ \\&&{9}(99+1)+ \\&&{7}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {82971}= &&\gray{8\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {8}+{2}+{9}+{7}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${82971}$ is divisible by $9$ if ${ 8}+{2}+{9}+{7}+{1}$ is divisible by $9$ Add the digits of ${82971}$ $ {8}+{2}+{9}+{7}+{1} = {27} $ If ${27}$ is divisible by $9$ , then ${82971}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${82971}$ must also be divisible by $9$.